46. Theqin the denominator is to be interpreted as|q|(so that the orbital radiusris a positive number).We interpret the given 10.0 MeV to be the kinetic energy of the electron. In order to make use of themc2value for the electron given in Table 38-3 (511 keV = 0.511 MeV) we write the classical kinetic energyformula asKclassical=12mv2=12(mc2)µv2c2¶=12(mc2)β2.(a) IfKclassical=10.0MeV, thenβ=r2Kclassicalmc2=r2(10.0MeV)0.511 MeV=6.256,which, of course, is impossible (see the Ultimate Speed subsection of§38-2). If we use this valueanyway, then the classical orbital radius formula yieldsr=mv|q|B=mβceB=(9.11×10−31kg)(6.256)(2.998×108m/s)(1.6×10−19C) (2.20 T)=4.85×10−3m.If, however, we use the correct value forβ(calculated in the next part) then the classical radiusformula would give about 0.77 mm.(b) Before using the relativistically correct orbital radius formula, we must compute
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.